This document contains a mathematics assignment on differentiation from Politeknik Manufaktur Negeri Bangka Belitung. It includes 15 practice problems finding derivatives of various functions. The problems are worked out step-by-step with the solutions. The assignment was prepared by 4 students in the Electronics Engineering program for their 2nd semester of the 2014/2015 academic year.

1. POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
1
TUGAS KALKULUS
( DIFFERENTIATION )
( Halaman 23-31 )
MATEMATIKA 2
Disusun Oleh :
Nama :
1. SIRILUS OKI SELPHADINATA
2. SITI FATIMAH
3. ODI BARKAH
4. ANDEKI
Prodi : Teknik Elektronika
Kelas : 1E A
Semester : 2 (Genap)
POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
Kawasan Industri Air Kantung Sungailiat, Bangka 33211
Telp. (0717) 93586, Fax. (0717) 93585
Email : polman@polman-babel.ac.id
Website : www.polman-babel.ac.id
TAHUN AJARAN 2014/2015
Latihan 5.1
1. 𝑓( 𝑥) = 2𝑥3

2. POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
2
𝑓′( 𝑥) = 2
𝑑
𝑑𝑥
( 𝑥3)
𝑓′( 𝑥) = 6𝑥2
2. 𝑔( 𝑥) =
𝑥100
25
𝑔′( 𝑥) =
𝑑
𝑑𝑥
(
𝑥100
25
)
𝑔′( 𝑥) = 100𝑥99
3. 𝑓( 𝑥) = 20𝑥
1
2
𝑓′( 𝑥) = 20
𝑑
𝑑𝑥
(𝑥
1
2)
𝑓′( 𝑥) = 10𝑥
−
1
2
4. 𝑦 = −16√ 𝑥
𝑦 = −16𝑥
1
2
𝑦′ = −16
𝑑
𝑑𝑥
(𝑥
1
2)
𝑦′ = −8𝑥
−
1
2
5. 𝑓( 𝑡) =
2𝑡
3
𝑓′( 𝑡) =
𝑑
𝑑𝑥
(
2𝑡
3
)
𝑓′( 𝑡) = 2𝑡
6. 𝑓( 𝑥) =
𝑥 𝜋
2𝜋
𝑓( 𝑥) = 2𝑥𝜋 𝜋
𝑓′( 𝑥) = 2
𝑑
𝑑𝑥
(𝑥𝜋 𝜋
)
𝑓′( 𝑥) = 2𝑥𝜋2𝜋
7. 𝑓( 𝑥) =
10
𝑥5
𝑓( 𝑥) = 10𝑥−5
𝑓′( 𝑥) = 10
𝑑
𝑑𝑥
(𝑥−5
)
𝑓′( 𝑥) = −50𝑥−6

3. POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
3
8. 𝑠( 𝑡) = 100𝑡0,6
𝑠′( 𝑡) = 100
𝑑
𝑑𝑥
(𝑡0,6
)
𝑠′( 𝑡) = 60𝑡−0,4
9. ℎ( 𝑠) = −25𝑠
1
2
ℎ′( 𝑠) = −25
𝑑
𝑑𝑥
(𝑠
1
2)
ℎ′( 𝑠) = −12,5 𝑠
−
1
2
10. 𝑓( 𝑥) =
1
4 √𝑥23
𝑓′( 𝑥) = 4𝑥
2
3
𝑓′( 𝑥) = 4
𝑑
𝑑𝑥
(𝑥
2
3)
𝑓′( 𝑥) =
8
3
𝑥
−
1
3
11. 𝑓′(3) 𝑤ℎ𝑒𝑛 𝑓( 𝑥) = 2𝑥3
𝑓′( 𝑥) = 2
𝑑
𝑑𝑥
( 𝑥3)
𝑓′( 𝑥) = 6𝑥2
𝑓′(3) = 6(3)2
𝑓′(3) = 54
12. 𝑔′( 𝑥) 𝑤ℎ𝑒𝑛 𝑔( 𝑥) =
𝑥100
25
𝑔′( 𝑥) =
𝑑
𝑑𝑥
(
𝑥100
25
)
𝑔′( 𝑥) = 100𝑥99
𝑔′(1) = 100(1)99
𝑔′(1) = 100
13. 𝑓′(81) 𝑤ℎ𝑒𝑛 𝑓( 𝑥) = 20𝑥
1
2
𝑓′( 𝑥) = 20
𝑑
𝑑𝑥
(𝑥
1
2)
𝑓′( 𝑥) = 10𝑥
−
1
2

4. POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
4
𝑓′(81) = 10(81)
−
1
2
𝑓′(81) =
10
9
14.
𝑑𝑦
𝑑𝑥
|25 𝑦 = −16√ 𝑥
𝑦 = −16𝑥
1
2
𝑦′ = −16
𝑑
𝑑𝑥
(𝑥
1
2)
𝑦′ = −8𝑥
−
1
2
𝑦′ = −8(25)
−
1
2
𝑦′
= −
8
5
15. 𝑓′(200) 𝑤ℎ𝑒𝑛 𝑓( 𝑡) =
2𝑡
3
𝑓′( 𝑡) =
𝑑
𝑑𝑥
(
2𝑡
3
)
𝑓′( 𝑡) = 2𝑡
𝑓′(200) = 2(200)
𝑓′(200) = 400
Latihan 5.2
1. 𝑓 ( 𝑥) = 𝑥7
+ 2𝑥10
𝑓′( 𝑥) =
𝑑
𝑑𝑥
( 𝑥7) +
𝑑
𝑑𝑥
(2𝑥10)

5. POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
5
𝑓′( 𝑥) = 7𝑥6
+ 20𝑥9
2. ℎ ( 𝑥) = 30 − 5𝑥2
ℎ′( 𝑥) =
𝑑
𝑑𝑥
(30) −
𝑑
𝑑𝑥
(5𝑥2)
ℎ′( 𝑥) = −10𝑥
3. 𝑔 ( 𝑥) = 𝑥100
− 40𝑥5
𝑔′( 𝑥) =
𝑑
𝑑𝑥
( 𝑥100) +
𝑑
𝑑𝑥
(40𝑥5)
𝑔′( 𝑥) = 100𝑥99
− 200𝑥4
4. 𝑐 ( 𝑥) = 1000 + 200𝑥 − 40𝑥2
𝑐′( 𝑥) =
𝑑
𝑑𝑥
(1000) +
𝑑
𝑑𝑥
(200𝑥) +
𝑑
𝑑𝑥
(40𝑥2
)
𝑐′( 𝑥) = 200 − 80𝑥
5. 𝑦 =
−15
𝑥
+ 25
𝑦′ =
𝑑
𝑑𝑥
(
−15
𝑥
) +
𝑑
𝑑𝑥
(25)
𝑦′
= −15𝑥−2
6. 𝑠 ( 𝑡) = 16𝑡2
−
2𝑡
3
+ 10
𝑠′( 𝑡) =
𝑑
𝑑𝑥
(16𝑡2) −
𝑑
𝑑𝑥
(
2𝑡
3
) +
𝑑
𝑑𝑥
(10)
𝑠′( 𝑡) = 32𝑡 − 6𝑡
7. 𝑔 ( 𝑥) =
𝑥100
25
− 20√ 𝑥
𝑔′ ( 𝑥) =
𝑑
𝑑𝑥
(
𝑥100
25
) −
𝑑
𝑑𝑥
(20√ 𝑥)
𝑔′( 𝑥) = 100𝑥99
− 20
1
2
8. 𝑦 = 12𝑥0,2
+ 0,45𝑥
𝑦′
=
𝑑
𝑑𝑥
(12𝑥0,2) +
𝑑
𝑑𝑥
(0,45𝑥)
𝑦′
= 2,4𝑥−0,8
+ 0,45
9. 𝑞 ( 𝑣) = 𝑣
1
2 + 7 − 15𝑣
1
2

6. POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
6
𝑞′( 𝑣) =
𝑑
𝑑𝑥
( 𝑣
1
2) +
𝑑
𝑑𝑥
(7) −
𝑑
𝑑𝑥
(15𝑣
1
2)
𝑞′ ( 𝑣) =
1
2
𝑣
3
2 − 7,5𝑣
3
2
10. 𝑓 ( 𝑥) =
5
2𝑥2 +
5
2𝑥−2 −
5
2
𝑓 ( 𝑥) =
𝑑
𝑑𝑥
(10𝑥−2
) +
𝑑
𝑑𝑥
(10𝑥2
) −
𝑑
𝑑𝑥
(
5
2
)
𝑓′( 𝑥) = 20𝑥−3
+ 20𝑥
11. ℎ′ (
1
2
) 𝑤ℎ𝑒𝑛 ℎ ( 𝑥) = 30 − 5𝑥2
ℎ′( 𝑥) =
𝑑
𝑑𝑥
(30) −
𝑑
𝑑𝑥
(5𝑥2
)
ℎ′( 𝑥) = −10𝑥
ℎ′( 𝑥) = −10 (
1
2
) = −5
12. 𝑐′(300) 𝑤ℎ𝑒𝑛 𝑐 ( 𝑥) = 1000 + 200𝑥 − 40𝑥2
𝑐′( 𝑥) =
𝑑
𝑑𝑥
(1000) +
𝑑
𝑑𝑥
(200𝑥) −
𝑑
𝑑𝑥
(40𝑥2
)
𝑐′( 𝑥) = 200 − 80𝑥
𝑐′(300) = 200 − 80(300) = −23800
13. 𝑠′(0) 𝑤ℎ𝑒𝑛 𝑠 ( 𝑡) = 16𝑡2
−
2𝑡
3
+ 10
𝑠′( 𝑡) =
𝑑
𝑑𝑥
(16𝑡2) −
𝑑
𝑑𝑥
(
2𝑡
3
) +
𝑑
𝑑𝑥
(10)
𝑠′( 𝑡) = 32𝑡 − 6𝑡
𝑠′(0) = 32(0) − 6(0) = 0
14. 𝑞′(32) 𝑤ℎ𝑒𝑛 𝑞 ( 𝑣) = 𝑣
1
2 + 7 − 15𝑣
1
2
𝑞′( 𝑣) =
𝑑
𝑑𝑥
( 𝑣
1
2) +
𝑑
𝑑𝑥
(7) −
𝑑
𝑑𝑥
(15𝑣
1
2)
𝑞′ ( 𝑣) =
1
2
𝑣
3
2 − 7,5𝑣
3
2
𝑞′ (32) =
1
2
(32)
3
2 − 7,5(32)
3
2 = −1267,14
15. 𝑓′(6) 𝑤ℎ𝑒𝑛 𝑓 ( 𝑥) =
5
2𝑥2 +
5
2𝑥−2 −
5
2

7. POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
7
𝑓 ( 𝑥) =
𝑑
𝑑𝑥
(10𝑥−2
) +
𝑑
𝑑𝑥
(10𝑥2
) −
𝑑
𝑑𝑥
(
5
2
)
𝑓′( 𝑥) = 20𝑥−3
+ 20𝑥
𝑓′( 𝑥) = 20(6)−3
+ 20(6) = 121,85
Latihan 5.3
1. 𝑓 ( 𝑥) = (2𝑥2
+ 3) + (2𝑥 − 3)
𝑓′( 𝑥) = (2𝑥2
+ 3)
𝑑
𝑑𝑥
(2𝑥 − 3) + (2𝑥 − 3)
𝑑
𝑑𝑥
(2𝑥2
+ 3)
𝑓′( 𝑥) = (2𝑥2
+ 3)(2) + (2𝑥 − 3)(4𝑥)
𝑓′( 𝑥) = 4𝑥2
+ 6 + 8𝑥2
− 12𝑥
𝑓′( 𝑥) = 12𝑥2
− 12𝑥 + 6
2. ℎ ( 𝑥) = (4𝑥2
+ 1) + (−𝑥2
+ 2𝑥 + 5)

8. POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
8
ℎ′ ( 𝑥) = (4𝑥2
+ 1)
𝑑
𝑑𝑥
(−𝑥2
+ 2𝑥 + 5) + (−𝑥2
+ 2𝑥 + 5)
𝑑
𝑑𝑥
(4𝑥2
+ 1)
ℎ′ ( 𝑥) = (4𝑥2
+ 1)(−2𝑥 + 2) + (−𝑥2
+ 2𝑥 + 5)(8𝑥)
ℎ′ ( 𝑥) = (−8𝑥3
+ 8𝑥2
− 2𝑥 + 2) + (−8𝑥3
+ 16𝑥2
+ 40𝑥)
ℎ′ ( 𝑥) = −16𝑥3
+ 24𝑥2
+ 38𝑥 + 2
3. 𝑔 ( 𝑥) = (𝑥2
− 5) + (
3
𝑥
)
𝑔′ ( 𝑥) = (𝑥2
− 5)(3𝑥−1
)
𝑔′ ( 𝑥) = (𝑥2
− 5)
𝑑
𝑑𝑥
(3𝑥−1
) + (3𝑥−1
)
𝑑
𝑑𝑥
(𝑥2
− 5)
𝑔′ ( 𝑥) = (𝑥2
− 5)(−3𝑥−2
) + (3𝑥−1
)(2𝑥)
𝑔′ ( 𝑥) = −3𝑥 + 15𝑥−2
+ 6𝑥
𝑔′ ( 𝑥) = 3𝑥 − 15𝑥−2
4. 𝑐 ( 𝑥) = (50 + 20𝑥)(100 − 2𝑥)
𝑐′ ( 𝑥) = (50 + 20𝑥)
𝑑
𝑑𝑥
(100 − 2𝑥) + (100 − 2𝑥)
𝑑
𝑑𝑥
(50 + 20𝑥)
𝑐′ ( 𝑥) = (50 + 20𝑥)(−2) + (100 − 2𝑥)(20)
𝑐′ ( 𝑥) = −100𝑥 − 40𝑥 + 2000 − 40𝑥
𝑐′ ( 𝑥) = −80𝑥 + 1900
5. 𝑦 = (
−15
√ 𝑥
+ 25)(√ 𝑥 + 5)
𝑦 = (−15𝑥
−
1
2 + 25)(𝑥
1
2 + 5)
𝑦′
= (−15𝑥
−
1
2 + 25)
𝑑
𝑑𝑥
( 𝑥
1
2 + 5) + ( 𝑥
1
2 + 5)
𝑑
𝑑𝑥
(−15𝑥
−
1
2 + 25)
𝑦′
= (−15𝑥
−
1
2 + 25)(
1
2
𝑥
−
1
2) + ( 𝑥
1
2 + 5) (7,5𝑥
−
3
2)
𝑦′
= 7,5𝑥−1
+22,5𝑥
−
1
2 + 7,5𝑥−1
+ 37,5𝑥
−
3
2
𝑦′
= 22,5𝑥
−
1
2 + 15𝑥−1
+ 37,5𝑥
−
3
2
6. 𝑠( 𝑡) = (4𝑡 −
1
2
) (5𝑡 +
3
4
)
𝑠′
(𝑡) = (4𝑡 −
1
2
)
𝑑
𝑑𝑥
(5𝑡 +
3
4
) + (5𝑡 +
3
4
)
𝑑
𝑑𝑥
(4𝑡 −
1
2
)
𝑠′( 𝑡) = (4𝑡 −
1
2
) (5) + (5𝑡 +
3
4
) (4)
𝑠′( 𝑡) = (20𝑡 −
5
2
) + (20𝑡 + 3)

9. POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
9
𝑠′( 𝑡) = 40𝑡 −
11
2
7. 𝑔( 𝑥) = (2𝑥3
+ 2𝑥2)(2√ 𝑥3
)
𝑔( 𝑥) = (2𝑥3
+ 2𝑥2) (2𝑥
1
3)
𝑔′( 𝑥) = (2𝑥3
+ 2𝑥2)
𝑑
𝑑𝑥
(2𝑥
1
3) + (2𝑥
1
3)
𝑑
𝑑𝑥
(2𝑥3
+ 2𝑥2)
𝑔′( 𝑥) = (2𝑥3
+ 2𝑥2)(
2
3
𝑥
−
2
3) + (2𝑥
1
3)(6𝑥2
+ 4𝑥)
𝑔′( 𝑥) = (
4
3
𝑥
7
3 +
4
3
𝑥
4
3) + (12𝑥
7
3 + 8𝑥
4
3)
𝑔′( 𝑥) =
40
3
𝑥
7
3 +
28
3
𝑥
4
3
8. 𝑓( 𝑥) = (
10
𝑥5
) (
𝑥3+1
5
)
𝑓( 𝑥) = (10𝑥−5)(3𝑥2
)
𝑓′
(𝑥) = (10𝑥−5)
𝑑
𝑑𝑥
(3𝑥2) + (3𝑥2)
𝑑
𝑑𝑥
(10𝑥−5)
𝑓′( 𝑥) = (10𝑥−5)(6𝑥) + (3𝑥2)(−50𝑥−6)
𝑓′( 𝑥) = 60𝑥−5
− 150𝑥−4
9. 𝑞( 𝑣) = (𝑣2
+ 7)(−5𝑣−2
+ 2)
𝑞′( 𝑣) = (𝑣2
+ 7)
𝑑
𝑑𝑥
(−5𝑣−2
+ 2) + (−5𝑣−2
+ 2)
𝑑
𝑑𝑥
(𝑣2
+ 7)
𝑞′( 𝑣) = (𝑣2
+ 7)(10𝑣−3
) + (−5𝑣−2
+ 2)(2𝑣)
𝑞′
(𝑣) = (10𝑣−1
+ 70𝑣−3
) + (−10𝑣−2
+ 4𝑣)
𝑞′( 𝑣) = 4𝑣 + 10𝑣−1
−10𝑣−2
+ 70𝑣−3
10. 𝑓( 𝑥) = (2𝑥3
+ 3)(3 − √𝑥23
)
𝑓( 𝑥) = (2𝑥3
+ 3) (3 − 𝑥
2
3)
𝑓′( 𝑥) = (2𝑥3
+ 3)
𝑑
𝑑𝑥
(3 − 𝑥
2
3) + (3 − 𝑥
2
3)
𝑑
𝑑𝑥
(2𝑥3
+ 3)
𝑓′( 𝑥) = (2𝑥3
+ 3)(
2
3
𝑥
−
1
3) + (3 − 𝑥
2
3)(6𝑥2
)
𝑓′( 𝑥)
= (
4
3
𝑥
8
3 + 2𝑥
−
1
3) + (18𝑥2
− 6
8
3)
𝑓′( 𝑥) =
22
3
𝑥
8
3 + 18𝑥2
+ 2𝑥
−
1
3

10. POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
10
11. 𝑓′(15) 𝑤ℎ𝑒𝑛 𝑓 ( 𝑥) = (2𝑥2
+ 3) + (2𝑥 − 3)
𝑓′( 𝑥) = (2𝑥2
+ 3)
𝑑
𝑑𝑥
(2𝑥 − 3) + (2𝑥 − 3)
𝑑
𝑑𝑥
(2𝑥2
+ 3)
𝑓′( 𝑥) = (2𝑥2
+ 3)(2) + (2𝑥 − 3)(4𝑥)
𝑓′( 𝑥) = 4𝑥2
+ 6 + 8𝑥2
− 12𝑥
𝑓′( 𝑥) = 12𝑥2
− 12𝑥 + 6
𝑓′(15) = 12(15)2
− 12(15) + 6 = 2526
12. 𝑔′
(10)𝑤ℎ𝑒𝑛 𝑔 ( 𝑥) = (𝑥2
− 5) + (
3
𝑥
)
𝑔′ ( 𝑥) = (𝑥2
− 5)(3𝑥−1
)
𝑔′ ( 𝑥) = (𝑥2
− 5)
𝑑
𝑑𝑥
(3𝑥−1
) + (3𝑥−1
)
𝑑
𝑑𝑥
(𝑥2
− 5)
𝑔′ ( 𝑥) = (𝑥2
− 5)(−3𝑥−2
) + (3𝑥−1
)(2𝑥)
𝑔′ ( 𝑥) = −3𝑥 + 15𝑥−2
+ 6𝑥
𝑔′ ( 𝑥) = 3𝑥 − 15𝑥−2
𝑔′ (10) = 3(10) − 15(10)−2
= 30,15
13. 𝑐′(150) 𝑤ℎ𝑤𝑛 𝑐 ( 𝑥) = (50 + 20𝑥)(100 − 2𝑥)
𝑐′ ( 𝑥) = (50 + 20𝑥)
𝑑
𝑑𝑥
(100 − 2𝑥) + (100 − 2𝑥)
𝑑
𝑑𝑥
(50 + 20𝑥)
𝑐′ ( 𝑥) = (50 + 20𝑥)(−2) + (100 − 2𝑥)(20)
𝑐′ ( 𝑥) = −100𝑥 − 40𝑥 + 2000 − 40𝑥
𝑐′ ( 𝑥) = −80𝑥 + 1900
𝑐′ (150) = −80(150) + 1900 = −10100
14.
𝑑𝑦
𝑑𝑥
| 𝑥 = 25| 𝑦 = (
−15
√ 𝑥
+ 25)(√ 𝑥 + 5)
𝑦 = (−15𝑥
−
1
2 + 25)(𝑥
1
2 + 5)
𝑦′
= (−15𝑥
−
1
2 + 25)
𝑑
𝑑𝑥
( 𝑥
1
2 + 5) + ( 𝑥
1
2 + 5)
𝑑
𝑑𝑥
(−15𝑥
−
1
2 + 25)
𝑦′
= (−15𝑥
−
1
2 + 25)(
1
2
𝑥
−
1
2) + ( 𝑥
1
2 + 5) (7,5𝑥
−
3
2)
𝑦′
= 7,5𝑥−1
+22,5𝑥
−
1
2 + 7,5𝑥−1
+ 37,5𝑥
−
3
2
𝑦′
= 22,5𝑥
−
1
2 + 15𝑥−1
+ 37,5𝑥
−
3
2

11. POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
11
15. 𝑓′(2) 𝑤ℎ𝑒𝑛 𝑓( 𝑥) = (
10
𝑥5
) (
𝑥3+1
5
)
𝑓( 𝑥) = (10𝑥−5)(3𝑥2
)
𝑓′
(𝑥) = (10𝑥−5)
𝑑
𝑑𝑥
(3𝑥2) + (3𝑥2)
𝑑
𝑑𝑥
(10𝑥−5)
𝑓′( 𝑥) = (10𝑥−5)(6𝑥) + (3𝑥2)(−50𝑥−6)
𝑓′( 𝑥) = 60𝑥−5
− 150𝑥−4
𝑓′(2) = 60(2)−5
− 150(2)−4
= −
147
16
Latihan 5.4
1. 𝑓( 𝑥) =
5𝑥+2
3𝑥−1
𝑓′( 𝑥) =
(3𝑥−1)
𝑑
𝑑𝑥
(5𝑥+2)−(5𝑥+2)
𝑑
𝑑𝑥
(3𝑥−1)
(3𝑥−1)2
𝑓′( 𝑥) =
(3𝑥−1)(5)−(5𝑥+2)(3)
(3𝑥−1)2
𝑓′
(𝑥) =
5𝑥−5−15𝑥+6
9𝑥2−6𝑥+1
𝑓′
(𝑥) =
−10𝑥+1
9𝑥2−6𝑥+1

12. POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
12
2. ℎ( 𝑥) =
4−5𝑥2
8𝑥
ℎ′( 𝑥) =
(8𝑥)
𝑑
𝑑𝑥
(4−5𝑥2)−(4−5𝑥2)
𝑑
𝑑𝑥
(8𝑥)
(8𝑥)2
ℎ′( 𝑥) =
(8𝑥)(−10𝑥)−(4−5𝑥2)(8)
64𝑥2
ℎ′( 𝑥) =
−80𝑥2−32−40𝑥2
64𝑥2
ℎ′( 𝑥) =
−120𝑥2−32
64𝑥2
3. 𝑔( 𝑥) =
5
√ 𝑥
𝑔′( 𝑥) =
(√ 𝑥)
𝑑
𝑑𝑥
(5)−(5)
𝑑
𝑑𝑥
(√ 𝑥)
(√ 𝑥)2
𝑔′( 𝑥) =
(√ 𝑥)(0)−(5)(𝑥
1
2)
(√ 𝑥)2
𝑔′
(𝑥) =
−(5)( 𝑥
1
2)
𝑥
𝑔′( 𝑥) =
5
2𝑥
3
2
4. 𝑓( 𝑥) =
3𝑥
3
2−1
2𝑥
1
2+6
𝑓′( 𝑥) =
(2𝑥
1
2+6)
𝑑
𝑑𝑥
(3𝑥
3
2−1)−(3𝑥
3
2−1)
𝑑
𝑑𝑥
(2𝑥
1
2+6)
(2𝑥
1
2+6)
2
𝑓′( 𝑥) =
(2𝑥
1
2+6)(
9
2
𝑥
1
2)−(3𝑥
3
2−1) (𝑥
−
1
2)
(2𝑥
1
2+6)
2
5. 𝑦 = −
15
𝑥
𝑦′ =
(−15)
𝑑
𝑑𝑥
(−𝑥)−(−𝑥)
𝑑
𝑑𝑥
(−15)
(−𝑥)2
𝑦′ =
(−15)(−1)−(1)(0)
𝑥2
𝑦′ =
15
𝑥2

13. POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
13
6. 𝑠(𝑡) =
2𝑡
3
2−3
4𝑡
1
2+6
𝑠′(𝑡) =
(4𝑡
1
2+6)
𝑑
𝑑𝑥
(2𝑡
3
2−3)−(2𝑡
3
2−3)
𝑑
𝑑𝑥
(4𝑡
1
2+6)
(4𝑡
1
2+6)2
𝑠′(𝑡) =
(4𝑡
1
2+6)(3𝑡
1
2)−(2𝑡
3
2−3)(2𝑡
−
1
2)
(4𝑡
1
2+6)2
𝑠′( 𝑡) =
12𝑡+18𝑡
1
2−4𝑡−6𝑡
−
1
2
16𝑡+48𝑡
1
2+36
7. 𝑔( 𝑥) =
𝑥100
𝑥−5+10
𝑔′( 𝑥) =
(𝑥−5+10)
𝑑
𝑑𝑥
(𝑥100)−(𝑥100)
𝑑
𝑑𝑥
(𝑥−5+10)
(𝑥−5+10)2
𝑔′( 𝑥) =
(𝑥−5+10)(100𝑥99)−(𝑥100) (−5𝑥−6)
(𝑥−5+10)2
𝑔′( 𝑥) =
100𝑥94+1000𝑥99+5𝑥94
𝑥−25+20𝑥−5+100
𝑔′( 𝑥) =
105𝑥94+1000𝑥99
𝑥−25+20𝑥−5+100
8. 𝑦 =
4−5𝑥3
8𝑥2−7
𝑦′ =
(8𝑥2−7)
𝑑
𝑑𝑥
(4−5𝑥3)−(4−5𝑥3)
𝑑
𝑑𝑥
(8𝑥2−7)
(8𝑥2−7)2
𝑦′ =
(8𝑥2−7)(−15𝑥2)−(4−5𝑥3)(16𝑥)
(8𝑥2−7)2
𝑦′ =
−120𝑥4−80𝑥3+105𝑥2−64𝑥
64𝑥4−112𝑥2+49
9. 𝑞(𝑣) =
𝑣3 +2
𝑣2−
1
𝑣2
𝑞(𝑣) =
𝑣3+2
𝑣2−𝑣−2
𝑞′(𝑣) =
( 𝑣2−𝑣−2)
𝑑
𝑑𝑥
( 𝑣3+2)−(𝑣3+2)
𝑑
𝑑𝑥
(𝑣2−𝑣−2)
(𝑣2−𝑣−2)2
𝑞′(𝑣) =
( 𝑣2−𝑣−2)(3𝑣2)−(𝑣3+2)(2𝑣+2𝑣−3)
𝑣4−2𝑣+𝑣−4
𝑞′(𝑣) =
3𝑣4−3𝑣−2𝑣4+2𝑣+4𝑣+4𝑣−3
𝑣4−2𝑣+𝑣−4

14. POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
14
𝑞′(𝑣) =
𝑣4+3𝑣+4𝑣−3
𝑣4−2𝑣+𝑣−4
10. 𝑓(𝑥) =
−4𝑥2
4
𝑥2+8
𝑓(𝑥) =
−4𝑥2
4𝑥−2+8
𝑓′(𝑥) =
(4𝑥−2+8)
𝑑
𝑑𝑥
(−4𝑥2)−(−4𝑥2)
𝑑
𝑑𝑥
(4𝑥−2+8)
(4𝑥−2+8)2
𝑓′(𝑥) =
(4𝑥−2+8)(−8𝑥)−(−4𝑥2)(−8𝑥−3)
(4𝑥−2+8)2
𝑓′(𝑥) =
−36𝑥−1−64𝑥−36𝑥−1
16𝑥−4+64𝑥−2+64
𝑓′(𝑥) =
−64𝑥−1−64𝑥
16𝑥−4+64𝑥−2+64
11. 𝑓′(25) 𝑤ℎ𝑒𝑛 𝑓( 𝑥) =
5𝑥+2
3𝑥−1
𝑓′( 𝑥) =
(3𝑥−1)
𝑑
𝑑𝑥
(5𝑥+2)−(5𝑥+2)
𝑑
𝑑𝑥
(3𝑥−1)
(3𝑥−1)2
𝑓′( 𝑥) =
(3𝑥−1)(5)−(5𝑥+2)(3)
(3𝑥−1)2
𝑓′
(𝑥) =
5𝑥−5−15𝑥+6
9𝑥2−6𝑥+1
𝑓′
(𝑥) =
−10𝑥+1
9𝑥2−6𝑥+1
𝑓′(25) =
−10(25)+1
9(25)2−6(25)+1
= −
249
5476
12. ℎ′(0,2) 𝑤ℎ𝑒𝑛 ℎ( 𝑥) =
4−5𝑥2
8𝑥
ℎ′( 𝑥) =
(8𝑥)
𝑑
𝑑𝑥
(4−5𝑥2)−(4−5𝑥2)
𝑑
𝑑𝑥
(8𝑥)
(8𝑥)2
ℎ′( 𝑥) =
(8𝑥)(−10𝑥)−(4−5𝑥2)(8)
64𝑥2
ℎ′( 𝑥) =
−80𝑥2−32−40𝑥2
64𝑥2
ℎ′( 𝑥) =
−120𝑥2−32
64𝑥2
ℎ′(0,2) =
−120(0,2)2−32
64(0,2)2 = −
36,8
2,56
13. 𝑔′(0,25) 𝑤ℎ𝑒𝑛 𝑔( 𝑥) =
5
√ 𝑥

15. POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
15
𝑔′( 𝑥) =
(√ 𝑥)
𝑑
𝑑𝑥
(5)−(5)
𝑑
𝑑𝑥
(√ 𝑥)
(√ 𝑥)2
𝑔′( 𝑥) =
(√ 𝑥)(0)−(5)(𝑥
1
2)
(√ 𝑥)2
𝑔′
(𝑥) =
−(5)( 𝑥
1
2)
𝑥
𝑔′( 𝑥) =
5
2𝑥
3
2
𝑔′(0,25) =
5
2(0,25)
3
2
= 20
14. 𝑦′(10) 𝑤ℎ𝑒𝑛 𝑦 = −
15
𝑥
𝑦′ =
(−15)
𝑑
𝑑𝑥
(−𝑥)−(−𝑥)
𝑑
𝑑𝑥
(−15)
(−𝑥)2
𝑦′ =
(−15)(−1)−(1)(0)
𝑥2
𝑦′ =
15
𝑥2
𝑦′
=
15
(10)2 =
15
100
15. 𝑔′(1) 𝑤ℎ𝑒𝑛 𝑔( 𝑥) =
𝑥100
𝑥−5+10
𝑔′( 𝑥) =
(𝑥−5+10)
𝑑
𝑑𝑥
(𝑥100)−(𝑥100)
𝑑
𝑑𝑥
(𝑥−5+10)
(𝑥−5+10)2
𝑔′( 𝑥) =
(𝑥−5+10)(100𝑥99)−(𝑥100) (−5𝑥−6)
(𝑥−5+10)2
𝑔′( 𝑥) =
100𝑥94+1000𝑥99+5𝑥94
𝑥−25+20𝑥−5+100
𝑔′( 𝑥) =
105𝑥94+1000𝑥99
𝑥−25+20𝑥−5+100
𝑔′(1) =
105(1)94+1000(1)99
(1)−25+20(1)−5+100
=
1105
121

16. POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
16
Latihan 5.5
1. 𝑓( 𝑥) = (3𝑥2
− 10)3
𝑢 = 3𝑥2
− 10 → 𝑑𝑢 𝑑𝑥 = 6𝑥⁄
𝑦 = 𝑢3
→ 𝑑𝑦 𝑑𝑢 = 3𝑢2
= 3(3𝑥2
− 10)2⁄
𝑑𝑦
𝑑𝑥
=
𝑑𝑢
𝑑𝑥
.
𝑑𝑦
𝑑𝑢
= 6𝑥. 3(3𝑥2
− 10)2
= 18𝑥(3𝑥2
− 10)2
2. 𝑔( 𝑥) = 40(3𝑥2
− 10)3
𝑢 = 3𝑥2
− 10 → 𝑑𝑢 𝑑𝑥 = 6𝑥⁄
𝑦 = 40𝑢3
→ 𝑑𝑦 𝑑𝑢 = 120𝑢2
= 120(3𝑥2
− 10)2⁄
𝑑𝑦
𝑑𝑥
=
𝑑𝑢
𝑑𝑥
.
𝑑𝑦
𝑑𝑢
= 6𝑥. 120(3𝑥2
− 10)2
= 720𝑥(3𝑥2
− 10)2
3. ℎ( 𝑥) = 10(3𝑥2
− 10)−3

17. POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
17
𝑢 = 3𝑥2
− 10 → 𝑑𝑢 𝑑𝑥 = 6𝑥⁄
𝑦 = 10𝑢−3
→ 𝑑𝑦 𝑑𝑢 = −30𝑢−4
= −30(3𝑥2
− 10)−4⁄
𝑑𝑦
𝑑𝑥
=
𝑑𝑢
𝑑𝑥
.
𝑑𝑦
𝑑𝑢
= 6𝑥. (−30)(3𝑥2
− 10)−4
= −180𝑥(3𝑥2
− 10)−4
4. ℎ( 𝑥) = (√ 𝑥 + 3)2
𝑢 = √ 𝑥 + 3 → 𝑥
1
2 + 3 → 𝑑𝑢 𝑑𝑥 =
1
2
𝑥
−
1
2⁄
𝑦 = 𝑢2
→ 𝑑𝑦 𝑑𝑢 = 2𝑢 = 2(√ 𝑥 + 3)⁄
𝑑𝑦
𝑑𝑥
=
𝑑𝑢
𝑑𝑥
.
𝑑𝑦
𝑑𝑢
=
1
2
𝑥
−
1
2 .2(√ 𝑥 + 3)
= 𝑥
−
1
2 .2(√ 𝑥 + 3)
5. 𝑓( 𝑣) = (
1
𝑣2 − 𝑣)
2
𝑢 =
1
𝑣2 − 𝑣 → 𝑣−2
− 𝑣 → 𝑑𝑢 𝑑𝑥 = −2𝑣−3⁄
𝑦 = 𝑢2
→ 𝑑𝑦 𝑑𝑢 = 2𝑢 = 2(
1
𝑣2 − 𝑣)⁄
𝑑𝑦
𝑑𝑥
=
𝑑𝑢
𝑑𝑥
.
𝑑𝑦
𝑑𝑢
= −2𝑣−3
.2(
1
𝑣2 − 𝑣)
= −
2
𝑣3 .2(
1
𝑣2 − 𝑣)
6. 𝑦 =
1
(𝑥2−8)3 → 𝑦 = (𝑥2
− 8)−3
𝑢 = 𝑥2
− 8 → 𝑑𝑢 𝑑𝑥 = 2𝑥⁄
𝑦 = 𝑢−2
→ 𝑑𝑦 𝑑𝑢 = −3𝑢−4
= −3(𝑥2
− 8)−4⁄
𝑑𝑦
𝑑𝑥
=
𝑑𝑢
𝑑𝑥
.
𝑑𝑦
𝑑𝑢
= 2𝑥 .−3(𝑥2
− 8)−4
= −6𝑥(𝑥2
− 8)−4
= −
6𝑥
(𝑥2−8)−4
7. 𝑦 = √2𝑥3 + 5𝑥 + 1

18. POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
18
𝑢 = 2𝑥3
+ 5𝑥 + 1 → 𝑑𝑢 𝑑𝑥 = 6𝑥2
+ 5⁄
𝑦 = √ 𝑢 → 𝑢
1
2 →
1
2
𝑢
−
1
2 → 𝑑𝑦 𝑑𝑢 =
1
2√ 𝑢
=
1
2√2𝑥3+5𝑥+1
⁄
𝑑𝑦
𝑑𝑥
=
𝑑𝑢
𝑑𝑥
.
𝑑𝑦
𝑑𝑢
= 6𝑥2
+ 5.
1
2√2𝑥3+5𝑥+1
8. 𝑠( 𝑡) = (2𝑡3
+ 5𝑡)
1
2
𝑢 = 2𝑡3
+ 5𝑡 → 𝑑𝑢 𝑑𝑥 = 6𝑡2
+ 5⁄
𝑦 = 𝑢
1
2 → 𝑑𝑦 𝑑𝑢 =
1
2
𝑢
−
1
2 =
1
2
(2𝑡3
+ 5𝑡)
−
1
2⁄
𝑑𝑦
𝑑𝑥
=
𝑑𝑢
𝑑𝑥
.
𝑑𝑦
𝑑𝑢
= 6𝑡2
+ 5.
1
2
(2𝑡3
+ 5𝑡)
−
1
2
9. 𝑓(𝑥) =
10
(2𝑥−6)5
𝑓(𝑥) = 10(2𝑥 − 6)−5
𝑢 = 2𝑥 − 6 → 𝑑𝑢 𝑑𝑥 = 2⁄
𝑦 = 10𝑢−5
→ 𝑑𝑦 𝑑𝑢 = −50𝑢−6
= −50(2𝑥 − 6)−6⁄
𝑑𝑦
𝑑𝑥
=
𝑑𝑢
𝑑𝑥
.
𝑑𝑦
𝑑𝑢
= 2 .−50(2𝑥 − 6)−6
= −100(2𝑥 − 6)−6
= −
100
(2𝑥−6)−6
10. 𝑐(𝑡) =
50
√15𝑡+120
𝑐(𝑡) = 50(15𝑡 + 120)
−
1
2
𝑢 = 15𝑡 + 120 → 𝑑𝑢 𝑑𝑥 = 15⁄
𝑦 = 50𝑢
−
1
2 → 𝑑𝑦 𝑑𝑢 = −25𝑢
−
3
2 = −25(15𝑡 + 120)
−
3
2⁄
𝑑𝑦
𝑑𝑥
=
𝑑𝑢
𝑑𝑥
.
𝑑𝑦
𝑑𝑢
= 15 .−25(15𝑡 + 120)
−
3
2
= −375(15𝑡 + 120)
−
3
2

19. POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
19
11. 𝑓′(10) 𝑤ℎ𝑒𝑛 𝑓( 𝑥) = (3𝑥2
− 10)3
𝑢 = 3𝑥2
− 10 → 𝑑𝑢 𝑑𝑥 = 6𝑥⁄
𝑦 = 𝑢3
→ 𝑑𝑦 𝑑𝑢 = 3𝑢2
= 3(3𝑥2
− 10)2⁄
𝑑𝑦
𝑑𝑥
=
𝑑𝑢
𝑑𝑥
.
𝑑𝑦
𝑑𝑢
= 6𝑥. 3(3𝑥2
− 10)2
= 18𝑥(3𝑥2
− 10)2
= 18(10)(3(10)2
− 10)2
= 15138000
12. ℎ′(3) 𝑤ℎ𝑒𝑛 ℎ( 𝑥) = 10(3𝑥2
− 10)−3
𝑢 = 3𝑥2
− 10 → 𝑑𝑢 𝑑𝑥 = 6𝑥⁄
𝑦 = 10𝑢−3
→ 𝑑𝑦 𝑑𝑢 = −30𝑢−4
= −30(3𝑥2
− 10)−4⁄
𝑑𝑦
𝑑𝑥
=
𝑑𝑢
𝑑𝑥
.
𝑑𝑦
𝑑𝑢
= 6𝑥. (−30)(3𝑥2
− 10)−4
= −180𝑥(3𝑥2
− 10)−4
= −180(3)(3(3)2
− 10)−4
= −
540
83521
13. 𝑓′(144) 𝑤ℎ𝑒𝑛 𝑓( 𝑥) = (√ 𝑥 + 3)2
𝑢 = √ 𝑥 + 3 → 𝑥
1
2 + 3 → 𝑑𝑢 𝑑𝑥 =
1
2
𝑥
−
1
2⁄
𝑦 = 𝑢2
→ 𝑑𝑦 𝑑𝑢 = 2𝑢 = 2(√ 𝑥 + 3)⁄
𝑑𝑦
𝑑𝑥
=
𝑑𝑢
𝑑𝑥
.
𝑑𝑦
𝑑𝑢
=
1
2
𝑥
−
1
2 .2(√ 𝑥 + 3)
= 𝑥
−
1
2 .2(√ 𝑥 + 3)
= (144)
−
1
2 .2(√144 + 3)
=
5
2
14. 𝑓′(2) 𝑤ℎ𝑒𝑛 𝑓( 𝑣) = (
1
𝑣2 − 𝑣)
2
𝑢 =
1
𝑣2 − 𝑣 → 𝑣−2
− 𝑣 → 𝑑𝑢 𝑑𝑥 = −2𝑣−3⁄
𝑦 = 𝑢2
→ 𝑑𝑦 𝑑𝑢 = 2𝑢 = 2(
1
𝑣2 − 𝑣)⁄

20. POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
20
𝑑𝑦
𝑑𝑥
=
𝑑𝑢
𝑑𝑥
.
𝑑𝑦
𝑑𝑢
= −2𝑣−3
.2(
1
𝑣2 − 𝑣)
= −
2
𝑣3 .2(
1
𝑣2 − 𝑣)
= −
2
(2)3 .2(
1
(2)2 − (2))
= −
17
8
15. 𝑦(4) 𝑤ℎ𝑒𝑛 𝑦 =
1
(𝑥2−8)3 → 𝑦 = (𝑥2
− 8)−3
𝑢 = 𝑥2
− 8 → 𝑑𝑢 𝑑𝑥 = 2𝑥⁄
𝑦 = 𝑢−2
→ 𝑑𝑦 𝑑𝑢 = −3𝑢−4
= −3(𝑥2
− 8)−4⁄
𝑑𝑦
𝑑𝑥
=
𝑑𝑢
𝑑𝑥
.
𝑑𝑦
𝑑𝑢
= 2𝑥 .−3(𝑥2
− 8)−4
= −6𝑥(𝑥2
− 8)−4
= −
6𝑥
(𝑥2−8)−4
= −
6(4)
((4)2−8)−4
= −98304

21. POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
21
Latihan 5.6
1. 𝑥2
𝑦 = 1
𝑑
𝑑𝑥
( 𝑥2
𝑦) =
𝑑
𝑑𝑥
(1)
2𝑥
𝑑𝑥
𝑑𝑥
. 𝑦 + 𝑥2
.
𝑑𝑦
𝑑𝑥
= 0
2𝑥𝑦 + 𝑥2 𝑑𝑦
𝑑𝑥
= 0
𝑑𝑦
𝑑𝑥
=
2𝑥𝑦
𝑥2
2. 𝑥𝑦3
= 3𝑥2
𝑦 + 5𝑦
3𝑥3
𝑦4
− 5𝑦 = 0
𝑑
𝑑𝑥
(3𝑥3
𝑦4
− 5𝑦) =
𝑑
𝑑𝑥
(0)
𝑑
𝑑𝑥
(3𝑥3
𝑦4
) −
𝑑𝑦
𝑑𝑥
(5𝑦) =
𝑑
𝑑𝑥
(0)
9𝑥2 𝑑𝑥
𝑑𝑥
. 𝑦4
+ 3𝑥3
.4𝑦3 𝑑𝑦
𝑑𝑥
= 0
9𝑥2
𝑦4
+ 3𝑥3
4𝑦3 𝑑𝑦
𝑑𝑥
= 0
𝑑𝑦
𝑑𝑥
=
9𝑥2 𝑦4
3𝑥34𝑦3
3. √ 𝑥 + √ 𝑦 = 25
𝑑
𝑑𝑥
( 𝑥
1
2 + 𝑦
1
2) =
𝑑
𝑑𝑥
(25)
𝑑
𝑑𝑥
( 𝑥
1
2) +
𝑑𝑦
𝑑𝑥
( 𝑦
1
2) =
𝑑
𝑑𝑥
(25)

22. POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
22
1
2
𝑥
−
1
2 +
1
2
𝑦
−
1
2
𝑑𝑦
𝑑𝑥
= 0
𝑑𝑦
𝑑𝑥
= −
1
2
𝑥
−
1
2
1
2
𝑦
−
1
2
4.
1
𝑥
+
1
𝑦
= 9
𝑑
𝑑𝑥
( 𝑥−1
+ 𝑦−1) =
𝑑
𝑑𝑥
(9)
𝑑
𝑑𝑥
( 𝑥−1) +
𝑑𝑦
𝑑𝑥
( 𝑦−1) =
𝑑
𝑑𝑥
(9)
−𝑥−2
− 𝑦−2 𝑑𝑦
𝑑𝑥
= 0
𝑑𝑦
𝑑𝑥
=
𝑥−2
−𝑦−2
5. 𝑥2
+ 𝑦2
= 16
𝑑
𝑑𝑥
(𝑥2
+ 𝑦2
) =
𝑑
𝑑𝑥
(16)
𝑑
𝑑𝑥
( 𝑥2) +
𝑑𝑦
𝑑𝑥
( 𝑦2) =
𝑑
𝑑𝑥
(16)
2𝑥 + 2𝑦
𝑑𝑦
𝑑𝑥
= 0
𝑑𝑦
𝑑𝑥
= −
2𝑥
2𝑦
6.
𝑑𝑦
𝑑𝑥
|(3,1) 𝑤ℎ𝑒𝑛 𝑥2
𝑦 = 1
𝑑
𝑑𝑥
( 𝑥2
𝑦) =
𝑑
𝑑𝑥
(1)
2𝑥
𝑑𝑥
𝑑𝑥
. 𝑦 + 𝑥2
.
𝑑𝑦
𝑑𝑥
= 0
2𝑥𝑦 + 𝑥2 𝑑𝑦
𝑑𝑥
= 0
𝑑𝑦
𝑑𝑥
=
2𝑥𝑦
𝑥2
𝑑𝑦
𝑑𝑥
=
2(3)(1)
(3)2 =
6
9
7.
𝑑𝑦
𝑑𝑥
|(5,2) 𝑤ℎ𝑒𝑛 𝑥𝑦3
= 3𝑥2
𝑦 + 5𝑦
3𝑥3
𝑦4
− 5𝑦 = 0
𝑑
𝑑𝑥
(3𝑥3
𝑦4
− 5𝑦) =
𝑑
𝑑𝑥
(0)

23. POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
23
𝑑
𝑑𝑥
(3𝑥3
𝑦4
) −
𝑑𝑦
𝑑𝑥
(5𝑦) =
𝑑
𝑑𝑥
(0)
9𝑥2 𝑑𝑥
𝑑𝑥
. 𝑦4
+ 3𝑥3
.4𝑦3 𝑑𝑦
𝑑𝑥
= 0
9𝑥2
𝑦4
+ 3𝑥3
4𝑦3 𝑑𝑦
𝑑𝑥
= 0
𝑑𝑦
𝑑𝑥
=
9𝑥2 𝑦4
3𝑥34𝑦3
8. √ 𝑥 + √ 𝑦 = 25
𝑑
𝑑𝑥
( 𝑥
1
2 + 𝑦
1
2) =
𝑑
𝑑𝑥
(25)
𝑑
𝑑𝑥
( 𝑥
1
2) +
𝑑𝑦
𝑑𝑥
( 𝑦
1
2) =
𝑑
𝑑𝑥
(25)
1
2
𝑥
−
1
2 +
1
2
𝑦
−
1
2
𝑑𝑦
𝑑𝑥
= 0
𝑑𝑦
𝑑𝑥
= −
1
2
𝑥
−
1
2
1
2
𝑦
−
1
2
𝑑𝑦
𝑑𝑥
= −
1
2
(4)−
1
2
1
2
(9)−
1
2
= −
1
4
1
6
9.
1
𝑥
+
1
𝑦
= 9
𝑑
𝑑𝑥
( 𝑥−1
+ 𝑦−1) =
𝑑
𝑑𝑥
(9)
𝑑
𝑑𝑥
( 𝑥−1) +
𝑑𝑦
𝑑𝑥
( 𝑦−1) =
𝑑
𝑑𝑥
(9)
−𝑥−2
− 𝑦−2 𝑑𝑦
𝑑𝑥
= 0
𝑑𝑦
𝑑𝑥
=
𝑥−2
−𝑦−2
𝑑𝑦
𝑑𝑥
= −
(5)−2
(10)−2 = −
1
25
1
100
= −4
10. 𝑥2
+ 𝑦2
= 16
𝑑
𝑑𝑥
(𝑥2
+ 𝑦2
) =
𝑑
𝑑𝑥
(16)
𝑑
𝑑𝑥
( 𝑥2) +
𝑑𝑦
𝑑𝑥
( 𝑦2) =
𝑑
𝑑𝑥
(16)
2𝑥 + 2𝑦
𝑑𝑦
𝑑𝑥
= 0

24. POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
24
𝑑𝑦
𝑑𝑥
= −
2𝑥
2𝑦
𝑑𝑦
𝑑𝑥
= −
2(2)
2(1)
= −2